What is the integral of (-2*x^4)/(2*x^3-3*x^2) ?

The integral of (-2*x^4)/(2*x^3-3*x^2) is -2((x^2)/4+3/4 x+9/8 ln|2x-3|)+C

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integral of (-2x^4)/(2x^3-3*x^2)

\int \frac{- 2 \cdot x ^{4}}{2 \cdot x ^{3} - 3 \cdot x ^{2}} d x

- 2 (\frac{x ^{2}}{4} + \frac{3}{4} x + \frac{9}{8} ln ∣ 2 x - 3 ∣) + C

Solution steps

\int \frac{- 2 x ^{4}}{2 x ^{3} - 3 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 \cdot \int \frac{x ^{4}}{2 x ^{3} - 3 x ^{2}} d x

Simplify

\frac{x ^{4}}{2 x ^{3} - 3 x ^{2}} : \frac{x ^{2}}{2 x - 3}

= - 2 \cdot \int \frac{x ^{2}}{2 x - 3} d x

Long division

\frac{x ^{2}}{2 x - 3} : \frac{x}{2} + \frac{3}{4} + \frac{9}{4 ( 2 x - 3 )}

= - 2 \cdot \int \frac{x}{2} + \frac{3}{4} + \frac{9}{4 ( 2 x - 3 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 2 (\int \frac{x}{2} d x + \int \frac{3}{4} d x + \int \frac{9}{4 ( 2 x - 3 )} d x)

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

\int \frac{3}{4} d x = \frac{3}{4} x

\int \frac{9}{4 ( 2 x - 3 )} d x = \frac{9}{8} ln ∣ 2 x - 3 ∣

= - 2 (\frac{x ^{2}}{4} + \frac{3}{4} x + \frac{9}{8} ln ∣ 2 x - 3 ∣)

Add a constant to the solution

= - 2 (\frac{x ^{2}}{4} + \frac{3}{4} x + \frac{9}{8} ln ∣ 2 x - 3 ∣) + C

\int \frac{x ^{4} + x - 25}{x ^{2} + 5} d x

\int x^{2} (1 - x)^{\frac{1}{2}} d x

\frac{d y}{d x} + 6 y = 0

\int_{2}^{3} x^{5} d x

f (x) = x^{2} - 2 x + 1, g (x) = - x + 3

What is the integral of (-2*x^4)/(2*x^3-3*x^2) ?
The integral of (-2*x^4)/(2*x^3-3*x^2) is -2((x^2)/4+3/4 x+9/8 ln|2x-3|)+C